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Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is

$- \frac{5 Gm}{r}$

$- \frac{6 Gm}{r}$

$- \frac{9 Gm}{r}$

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Explanation

$(3) L e t g r a v i t a t i o n a l f i e l d i s z e r o a t x d i s \tan c e f r o m m a s s m . \Rightarrow \frac{G m}{x^{2}} = \frac{4 G m}{{(r - x)}^{2}} \Rightarrow {(r - x)}^{2} = 4 x^{2} \Rightarrow r - x = 2 x \Rightarrow x = \frac{r}{3} P o t e n t i a l a t t h a t p o i n t : V = \frac{- G m}{x} - \frac{4 G m}{r - x} V = \frac{- G m}{\frac{r}{3}} - \frac{4 G m}{\frac{2 r}{3}} V = \frac{- 9 G m}{r}$

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The satellite of mass m orbiting around the earth in a circular orbit with a velocity v. The total energy will be:

$\frac{3}{4} {mv}^{2}$

$\frac{1}{2} {mv}^{2}$

$- \frac{1}{2} {mv}^{2}$

${mv}^{2}$

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Explanation

$E = - K = - \frac{m v^{2}}{2}$

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Magnitude of potential energy (U) and time period (T) of a satellite are related to each other as:

$T^{2} α \frac{1}{U^{3}}$

$T α \frac{1}{U^{3}}$

$T^{2} α U^{3}$

$T^{2} α \frac{1}{U^{2}}$

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Explanation

$|U| = \frac{G M m}{r} o r r α \frac{1}{U} T = \frac{2 πr}{V} = \frac{2 {πr}^{3 / 2}}{\sqrt{G m}} T^{2} α r^{3} α \frac{1}{U^{3}}$

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A projectile fired vertically upwards with a speed v escapes from the earth. If it is to be fired at 45 $°$ to the horizontal, what should be its speed so that it escapes from the earth?

$\frac{v}{\sqrt{2}}$

$\sqrt{2} v$

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Explanation

Escape velocity is independent of the angle of projection.

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Kepler's second law regarding constancy of areal velocity of a planet is a consequence of law of conservation of

Energy

Linear momentum

Angular momentum

Mass

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Explanation

$A n g u l a r m o m e n t u m = 2 \times m a s s \times a r e a l s p e e d$

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A body of super dense material with mass twice the mass of the earth but size very small compared to size of the earth starts from rest from h<<R above the Earth's surface. It reaches earth in time t:

$t = \sqrt{\frac{h}{g}}$

$t = \sqrt{\frac{2 h}{g}}$

$t = \sqrt{\frac{2 h}{3 g}}$

$t = \sqrt{\frac{4 h}{3 g}}$

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Explanation

$F o r c e b e t w e e n t h e b o d y a n d e a r t h = \frac{2 G M^{2}}{R^{2}} A c c e l e r a t i o n o f t h e b o d y = \frac{F_{g}}{2 M} = \frac{2 G M^{2}}{2 M R^{2}} = \frac{G M}{R^{2}} = g A c c e l e r a t i o n o f t h e e a r t h = \frac{F_{g}}{M} = \frac{2 G M^{2}}{M R^{2}} = \frac{2 G M}{R^{2}} = 2 g A c c e l e r a t i o n o f t h e b o d y w . r . t e a r t h = g - (- 2 g) = 3 g T i m e = \sqrt{\frac{2 h}{a}} = \sqrt{\frac{2 h}{3 g}}$

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A thin rod of length L is bent to form a semicircle. The mass of the rod is M. The gravitational potential at the centre of the circle is :

$- \frac{GM}{L}$

$- \frac{GM}{2 πL}$

$- \frac{πGM}{2 L}$

$- \frac{πGM}{L}$

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Explanation

$(4) As the rod is bent in the form of a semicircle, πR = L Radius of the circle, R = \frac{L}{π} V = - \frac{GM}{R} = - \frac{GM}{\frac{L}{π}} = - \frac{πGM}{L}$

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A point P lies on the axis of a ring of mass M and radius 'a' at a distance 'a' from its centre C. A small particle starts from P and reaches C under gravitational attraction. Its speed at C will be :

$\sqrt{\frac{2 GM}{a}}$

$\sqrt{\frac{2 GM}{a} (1 - \frac{1}{\sqrt{2}})}$

$\sqrt{\frac{2 GM}{a} (\sqrt{2} - 1)}$

zero

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Explanation

$(2) K_{p} + U_{p} = K_{c} + U_{c} 0 - \frac{GMm}{\sqrt{a^{2} + a^{2}}} = \frac{1}{2} {mv}^{2} - \frac{GMm}{a} v^{2} = 2 \times (- \frac{GMm}{\sqrt{2} a} + \frac{GMm}{a}) = \frac{2 GMm}{a} (- \frac{1}{\sqrt{2}} + 1) v = \sqrt{\frac{2 GMm}{a} (1 - \frac{1}{\sqrt{2}})}$

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Weightlessness experienced while orbiting the earth in space-ship, is the result of

Inertia

Acceleration

Zero gravity

Free fall towards earth

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Explanation

(d)

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The escape velocity for a rocket from earth is 11.2 km/sec. Its value on a planet where acceleration due to gravity is double that on the earth and diameter of the planet is twice that of earth will be in km/sec

11.2

5.6

22.4

53.6

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Explanation

(c) $\frac{v_{p}}{v_{e}} = \sqrt{\frac{g_{p}}{g_{e}} \times \frac{R_{p}}{R_{e}}} = \sqrt{2 \times 2} = 2 \Rightarrow v_{p} = 2 \times v_{e} = 2 \times 11.2 = 22.4 k m / s$

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Yes. You can attempt every NEET question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.